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Show that if a Markov chain is irreducible and has a state $s_i$ such that $P_{ii}>0$, then it is also aperiodic.

Proof: Let $X=(X_0, X_1, X_2, \dots)$ be an irreducible Markov chain with a state $s_i$ such that $P_{ii}>0$. Recall a Markov chain $(X_0,X_1,\dots)$ with state space $S=\{s_1,s_2,\dots\}$ and transition matrix $P$ is said to be \textit{irreducible} if for all $s_i, s_j\in S$ we have that $s_i\iff s_j$. We need to show that $X$ is also aperiodic.

A Markov chain is said to be aperiodic if all its states are aperiodic. The period $$d(s_i)=\gcd\{n\geq 1: (P^n)_{i,i}>0\}$$ of a state $s_i\in S$. If $d(s_i)=1$, then we say that the state $s_i$ is aperiodic.

Where do I go from here?

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  • $\begingroup$ Periodicity is a class property, does that help you? See for example point 2 of this website. $\endgroup$
    – Ritz
    Commented May 24, 2016 at 14:25
  • $\begingroup$ Not really, but will number 2 from the site seems like something I can use. Do you agree? $\endgroup$ Commented May 24, 2016 at 14:31

2 Answers 2

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Since $X$ is irreducible, for any state $j$ there exist positive integers $n,n'$ such that $P_{ij}^n>0$ and $P_{ji}^{n'}>0$. Since $P_{ii}>0$, it follows that $P_{ii}^m>0$ for all positive integers $m$, and hence $$P_{jj}^{n+n'+m}\geqslant P_{ji}^{n'}P_{ii}^mP_{ij}^n>0. $$ This implies that the period of state $j$ is $1$. Since $j$ was arbitrary, we conclude that $X$ is aperiodic.

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Let $m$ and $n$ be such that $P_{ij}^m \cdot P_{ji}^n > 0$. Then

\begin{align*} P_{jj}^{n+m} &\geq P_{ji}^n \cdot P_{ij}^m > 0 \\ P_{jj}^{n+m+1} &\geq P_{ji}^n \cdot P_{ii} \cdot P_{ij}^m >0. \end{align*}

Now \begin{align*} \gcd(n+m+1,n+m) &= \gcd(n+m+1 -(n+m),n+m) \\ &= \gcd(1,n+m) \\ &= 1, \end{align*} i.e., d($s_j) = 1$. Since $j$ was arbitrary, this shows that the markov chain $X$ is aperiodic.

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